Operating Systems CS 370 s1

CPU Scheduling 1Operating Systems CS 370 CPU SchedulingText: Chapter 5 Operating Systems Concepts with Java, 7th Ed., Silberschatz, Galvin, GagneObjectives: During this class, the student shall be able to: . Define I/O-bound, CPU-bound and race condition. . Define the functions of the short-term, long-term scheduler. . Describe the difference between preemptive and nonpreemptive schedulers. . Define and be able to calculate: CPU utilization, throughput, wait time, response time, and turnaround time. . Be able to calculate the order and timing of processes through the CPU using the following algorithms: FCFS, RR, Priority, Shortest Job First, Shortest Remaining Time, Multilevel Feedback Queue. . Define the difference between SMP and asymmetric multiprocessing as it relates to operating systems. . Define processor affinity and load balancing, and why each is important.Time Allocation: Class time will be allocated as follows: Intro 1/2 hour Scheduling Criteria 1/2 hour Scheduling Algorithms 2 hours TOTAL: 3 hours CPU Scheduling 2CPU SchedulingIntroductionProgram execution is characterized by alternating sequences of: . CPU burst . I/O burstTwo types of programs: . I/O-bound: short CPU bursts . CPU-bound: long CPU bursts. . Goal: Balancing I/O-bound and CPU-bound jobs.CPU Scheduler . Long Term Scheduling: Decides when to add to the pool of processes to be executed.  Controls degree of multiprogramming.  When a job terminates, scheduler may decide to add one or more jobs.  Batch jobs: Spooled on disk until ready to process  Interactive: User logs in O.S. must accept new process. . Medium Term Scheduling: Decides when to add to the number of processes that are partially or fully in main memory.  Who wins memory?  Global scope memory allocation impacts decision. . Short Term Scheduler: Decides which available process will be executed next by the processor.Short Term Scheduler: . When CPU becomes idle, OS selects a process in ready queue to be executed. . The Ready queue holds PCBs.Short Term Scheduling functions include: . Enqueuer function: Adds PCB to appropriate Ready Queue (possibly after deciding priority). . Dispatcher function: Selects process for execution and passes control to it. Includes:  Switches context  Switches to user mode  Jumps to proper location in user program. CPU Scheduling 3CPU reschedules processes: 1. When context switches from running state to blocked state (e.g. I/O request) or yields CPU. 2. When process terminates. 3. When context switches from running state to ready state (e.g. timer interrupt) 4. When context switches from blocked state to ready state (e.g. I/O completion)Two types of scheduling: Nonpreemptive: Scheduling only takes place during 1 & 2 above. . The process retains the CPU until it releases the CPU by terminating or by blocking. . Used by Microsoft Windows 3.X Preemptive: Scheduling may take place during all four event types. . Can preempt on: clock or I/O interrupt, OS call, signal. . Advantage: Prevents any process from monopolizing the processor. . Disadvantage: More overhead due to context switching. . Used by Microsoft Windows 95 and later, UNIX.Race Condition Problem: If two processes share data, one may be in midst of updating the data when it is preempted and second process is run  Problem: Single processor, preemptive OS  Problem: Multiprocessor, preemptive or non-preemptive OS  Solution: Semaphores CPU Scheduling 4Scheduling CriteriaStatistics to measure successfulness of CPU scheduling algorithms: . Note: 1 Second = 1000ms (milliseconds) . Time_to_Process = Wait_Time + Service_TimeSystem Oriented Statistics . CPU Utilization: % of time CPU is busy processing programs.  %Utilization = Arrival_Rate * Service_Time  In 1 avg. second 7 processes arrive, with service time = 100 ms each: Utilization = 7*.1 = 70%  In real world, it should range from 40% to 90%. . Throughput: The number of processes that are completed per time unit.  Assume processing times: Process 1: 300 ms; Process 2: 400ms; Process 3: 300ms.  Throughput = 3 processes in 1 second. . Wait Time: Sum of periods spent waiting in the ready queue.  Wait Time = EndTime – ArrivalTime – ServiceTimeUser Oriented Statistics . Turnaround Time: The interval from time of submission to time of completion.  Used also with batch programs.  Includes non-CPU times: waiting to get into memory, waiting in ready queue, doing I/O.  TurnaroundTime = EndTime – ArrivalTime . Response Time: The interval from submission of a request until first response is produced.  Used with interactive programs.  Does not include time it takes to output the response (to the output device).  2 Seconds or less desirable.Other concerns: . Predictability: Response time does not vary widely. . Optimize average, but guarantee all users get good service. CPU Scheduling 5Scheduling AlgorithmsFirst Come First Serve = FCFS or FIFOFCFS: Processes that request the CPU first are allocated the CPU first. . Implemented with a FIFO queue of PCBs. . NonpreemptiveProcess Arrive Time Burst Time P1 0 24 P2 0 3 P3 0 3P1 P2 P3 |------|------|------| 0 24 27 30Average wait time: (0+24+27)/3 = 17 ms Average turnaround time: (24+27+30)/3 = 81/3 = 27 msWhat happens if the processes arrive in order P2, P3, P1? How does that impact Wait, Turnaround time?Conclusion: . CPU-bound processes may get and hold (hog) the CPU. . Varied and poor response for I/O-bound processes. . Poor use of I/O devices. CPU Scheduling 6Shortest Job FirstShortest Job First: CPU is assigned to the process that has the smallest expected CPU burst. . Examines the length of the next CPU-burst, not the total process length. . Nonpreemptive . Used in long-term scheduling . Problem: Must know length of next CPU request.Process Arrive Time Burst Time P1 0 6 P2 0 8 P3 0 7 P4 0 3P4 P1 P3 P2 |------|------|------|------| 0 3 9 16 24Average wait time: (0+3+9+16)/4 = 7ms. Average turnaround time: (3+9+16+24)/4 = 13ms.Conclusion: . Problems:  Starvation: Longer processes may never get time.  No preemption: Short processes must wait for a CPU-bound process.How to predict length: . Batch: Use process time limit requested when submitting job. . Interactive: Take running average of burst length and use as predictor.How to calculate running average length? S(n+1) = pt(n) + (1-p)S(n) where S(n) = running average at time n t(n) = burst length at time n p = weight factor, where 0 Given p = 0.5 S(n+1) t(n) S(n) 8 6 10 6 4 8 6 6 6 5 4 6 CPU Scheduling 7Shortest Remaining Time FirstShortest-Remaining-Time-First: Preempts current process if another process would complete sooner. . Always selects the process which will complete the soonest (like Shortest-Process- First) . Preemptive.Process Arrive Time Burst Time P1 0 8 P2 1 4 P3 2 9 P4 3 5P1 P2 P4 P1 P3 |--|------|------|------|------| 0 1 5 10 17 26Average wait time: ((10-1) + (1-1) + (17-2) + (5-3)) / 4 = 26/4 = 6.5ms Average turnaround time: (17 + (5-1) + (10-3) + (26-2)) / 4 = 13msConclusion: . Advantage: A short job is given immediate preference to a running longer job. CPU Scheduling 8Priority SchedulingPriority Scheduling: High priority processes scheduled before low priority processes. . A priority is associated with each process. . Priority defined by: (E.g.) Foreground, Background, Real Time, Nice . PreemptivePriorities can be defined: . Number of priorities may range. Assume 0-1024:  High priority may be priority 0 or 1024, depending on system. . External or Internal Priorities: Is priority determined external or internal to the O.S?  External: Importance of process, $ paid for computer use, department priority.  Internal: Time limits, ratio of average I/O burst to average CPU burst, Aging.Assume lower number is higher priority: Process Arrive Time Burst Time Priority P1 0 10 3 P2 0 1 1 P3 1 2 3 P4 1 1 4 P5 1 5 2P2 P5 P1 P3 P4 |----|------|------|------|----| 0 1 6 16 18 19Wait Times:  Priority 1: 0ms  Priority 2: 0ms  Priority 3: (6 + (16-1)) / 2 = 10.5ms  Priority 4: 18-1 = 17msHow does the above change if P6 arrives at time 8 with a priority of 1 and a burst time of 3?Conclusion: . Problems: Starvation or indefinite blocking. . Solution: Internal Priority  Aging: Raise priority for processes that have waited for a long time.  Example: Raise priority by one when process has waited 1 minute.  Eventually even lowest priority process becomes the highest priority process. CPU Scheduling 9Round RobinRound Robin: Every process gets a turn = one time quantum at CPU. . Timer interrupt causes scheduler to context switch. . Time quantum = time slice: 10-100ms long. . Preemptive . Ready queue is circular queue.Assume time quantum is 4 ms long. Process Arrive Time Burst Time P1 0 24 P2 0 3 P3 0 6P1 P2 P3 P1 P3 P1 P1 P1 P1 |------|------|------|------|------|------|------|------|------| 0 4 7 11 15 17 21 25 29 33Average Wait Time: (4+9+11)/3 = 8 msConclusion: Performance depends heavily on size of time quantum. . Optimum: 80% of CPU bursts should be < time quantum. . Optimum: Time quantum should be large with respect to context switch time.  If context switch is approx. 10% of time quantum, then about 10% of CPU time is spent in context switch. . Advantage: Performs reasonably well with short-term jobs. . Disadvantage: Additional overhead required due to additional context switching. . Disadvantage: I/O bound processes rarely use entire timeslice before blocking due to I/O.  Favors CPU-bound processes somewhat. CPU Scheduling 10Multi-level Feedback QueueMultilevel Feedback Queue: Processes hogging CPU sink to lower priority levels. 1. All processes start at Request Queue 0 (highest priority). 2. When timeslice completed put in Request Queue 1. 3. When timeslice completed put in Request Queue 2. 4. When timeslice completed put in RQ ...OS Implementation: . Operates strictly on priority:  Schedules processes in RQ 0 until it is empty, then processes RQ1... . All queues processed in FCFS order, except lowest priority queue that operates in RR. . Lower priority queues may get increasing time quantums. . Problem: Low priority queues may starve  Solution: Allow processes that have waited a certain length of time to rise back up the priority queues.Assume RQ0 time quantum = 4, and RQ doubles the time quantum. Process Arrive Time Burst Time P1 0 24 P2 0 3 P3 6 3 P4 6 8P1 P2 P3 P4 P1 P4 P1 |------|------|------|------|------|------|------| 0 4 7 10 14 22 26 38Conclusion: . Advantage: General CPU scheduling solution . Disadvantage: Complex CPU Scheduling 11MultiprocessingSymmetric MultiProcessing (SMP): Each processor runs an OS scheduler. Asymmetric MultiProcessing: One processor runs the scheduler and assigns jobs to the other processorsLoad Balancing: Attempts to equalize the load between all processors.  This can be accomplished by migrating processes between processors.  Problem: A process can already be in cache and repopulating the new cache may take time.  Solution: Processor Affinity: Assign a process to the processor where its cache is already loadedReal Systems Linux Three scheduling classes:  FIFO: FIFO real-time threads (preemptive by priority)  RR: Round-robin real-time threads (preemptive by timeslice, scheduled by priority)  Other: Non-real time threads (run if no real-time threads) o Non-real-time: priorities 100-139, default 120 o Active queue list and expired queue list Scheduling Algorithm:  Round-Robin within Priority: Time slices range between 10-200 ms, higher priority -> larger time slices  Priority altered to favor I/O-bound tasks over CPU-bound tasks; high-sleep rate I/O bound highest priority.Windows  Fully preemptive: Higher-numbered priority process preempts lower priority process  32 Ready Queues  16-31: Real-time queues: Priority is fixed  1-15: Variable level queues: For normal applications  Priority is lowered if timeslice completes or is a background window.  Priority is raised if I/O completes (keyboard raised lots, disk some)  0: Memory management  Idle Thread: Spare time or idle time task runs when nothing else runsJava Thread Scheduling  JVM uses Priority-based scheduling, but may or may not be Preemptive  May or may not use timeslicing (not checked recently)  If timeslicing not used, a thread may yield control using Thread.yield()  Priorities range from 1 to 10, where 1 is low and 10 is high. CPU Scheduling 12 Priority can be changed using setPriority() method.  The JVM never alters the priority of a thread. CPU Scheduling 13Algorithm Evaluation . Deterministic Modeling: Use a fixed scenario of processes/threads arriving, as in scheduling algorithm examples above. . Queuing Models: Uses mathematical (probabilistic) models to analyze given arrival rates, service times, (and potentially number of processors) to derive average wait time and turnaround time. . Simulation: Randomly generates job arrivals with random service times, and simulates the processing of these random jobs over a sufficiently long duration of time to generate average wait time and turnaround times. Queuing Statistics . Inter-arrival Time: The average timer between incoming requests . Arrival Rate: 1/InterArrivalTime = The number of requests that arrive on average in a specified time unit . Service Time: The average time it takes to service a request . Service Rate: 1/ServiceTime = The number of requests that can be handled in a specified time unit. . Offered Rate: ArrivalRate/ServiceRate = ServiceTime/InterArrivalTime = The average number of requests being handled in a system, if no requests are discarded. CPU Scheduling 14Exercise Assume the following conditions (where low numbers are higher priority). For each method, draw a timeline and complete the per-job and total Wait Time and Turnaround Time.Shortest Job First (No Preemption) Job Arrival Service Turnaround Wait Time # Time Time Time 1 0 22 0 43 3 14 5 85 6 36 11 3Total /AvgShortest Remaining Time First (Preemption) Job Arrival Service Turnaround Wait Time # Time Time Time 1 0 22 0 43 3 14 5 85 6 36 11 3Total /Avg CPU Scheduling 15Priority (Preemptive) Lower numbers = higher priority. Job Arrival Priority Service Turnaround Wait Time # Time Time Time 1 0 10 22 0 10 43 3 3 14 5 10 85 6 3 36 11 10 3Total /AvgRound Robin (Timeslice = 3 ms) Job Arrival Service Turnaroun Wait Time # Time Time d Time 1 0 22 0 43 3 14 5 85 6 36 11 3Total /Avg

Operating Systems CS 370 s1

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